In his paper "Comparing analytic assemby maps", J. Roe considers a proper and cocompact action of a countable group $\Gamma$ on a metric space $X$. He constructs the Hilbert $C^*_r(\Gamma)$-module $L^2_\Gamma(X)$ as the simultaneous completion of the $\mathbb C\Gamma$-module $C_c(X)$ under the $\mathbb C\Gamma$-valued inner product given by the formula $$ \langle\psi\mid\psi'\rangle_\Gamma=\sum_{\gamma\in\Gamma}\langle\psi\cdot g\mid\psi'\rangle\,\gamma. $$ He shows (Lemmas 2.1-3) that the C$^*$-algebra of $C^*_r(\Gamma)$-compact operators $\mathbb K_{C^*_r(\Gamma)}(L^2_\Gamma(X))$ is isomorphic to $C^*_\Gamma(X)$, the $\Gamma$-invariant uniform Roe algebra.
All of this is fine, but there seems to be a problem with the following claims, and this is what my question is about.
Roe claims that $\mathbb K_{C^*(\Gamma)}(L^2_\Gamma(X))$ (and hence $C^*_\Gamma(X)$) is Morita equivalent to $C^*_r(\Gamma)$. I fail to see this and my question is: Is this true? If so, how to prove it? If not, how to disprove it?
Looking at Roe's arguments, he deduces the above statement from the more general claim that for any (countably generated) Hilbert $C^*_r(\Gamma)$-module $E$, $\mathbb K_{C^*(\Gamma)}(E)$ is Morita equivalent to $C^*_r(\Gamma)$. What is true is that it is Morita equivalent to $\langle E\mid E\rangle$ (with $E$ setting up the equivalence), that is immediate. The general statement that Roe makes however seems to be false unless $E$ is (right-) full. I find it hard to believe that it is true in general.
Concerning the particular case of $E=L^2_\Gamma(X)$, it would be great see that this is in fact a full HIlbert $C^*_r(\Gamma)$-module. I fail to see this, however. One computes in case of the above example: $$ \langle\delta_m\mid\delta_n\rangle=(m-n,+)+(m+n,-), $$ so the impression is that perhaps the closed ideal generator by the inner products is not the entire algebra.